A hot-air balloonist, rising vertically with a constant velocity of magnitude v

Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s , releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground (Figure 1) . After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive. Part A Compute the position of the sandbag at a time 0.250 s after its release. y = 40.9 m SubmitMy AnswersGive Up Correct Significant Figures Feedback: Your answer 41.2 m was either rounded differently or used a different number of significant figures than required for this part. Part B Compute the velocity of the sandbag at a time 0.250 s after its release. v = m/s SubmitMy AnswersGive Up Part C Compute the position of the sandbag at a time 1.35 s after its release. y = m SubmitMy AnswersGive Up Part D Compute the velocity of the sandbag at a time 1.35 s after its release.

Explanation / Answer

initial height=+40 m

initial velocity=+5 m/s

using newton equation

disance travelled in 0.25

y=5*0.25-0.5*9.8*0,25^2=0.94375 m

position=40+0.94375=+40.94 m

(b) velocity=5-(9.8*0.25)=+2.55 m/s

(c) at t=1.35

position=40-2.18=37.82 m

(d) velocity=5-(1.35*9.8)=-8.23 m/s

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